Physics Asked on March 9, 2021
For simplicity, consider the one dimensional Poisson equation expressed in Cartesian coordinates: $$frac{d^2V}{dx^2}=-frac{rho}{epsilon_0}.$$ Now, we observe in this case that the charge density is not unique. For a potential of $V=2x+1$ and a potential of 5, we get the same charge distribution. What does is physically mean? Where am I going wrong?
One problem is the lack of dimensions. (You have one). Let's skip the potential and work with a field, normally:
$$ {bf nabla E} =rho / epsilon_0 $$
In 3 dimensions, the free space field falls off as $1/r^2 (and V propto 1/r)$. You can imagine conserved flux line spreading out in space as the pass through a surface of radius $propto r^2$.
In 2 dimensions you expect the field to fall off as $1/r$, with $V propto ln r$, again you have conserved lines of flux spreading out.
Now you have chosen 1 dimension, so any amount of flux can't spread out. A point charge any where leads to a constant field pointing towards (or away from) it, and it never falls off. That means a point charge anywhere leads to a linear ramp on $V(x)$--this seems counter intuitive, that a distance charge can lead to a linear diverging potential, but it's only because there is only 1 dimension working.
As mentioned in other answers, this term is fixed with boundary conditions. If there is a point charge at $x_0$, then
$$ V(x) propto |x-x_0| $$
Of course, the constant term is meaningless in any number of dimensions.
Correct answer by JEB on March 9, 2021
The very first thing is the potential is a relative quantity, just like the gravity potential, you can only talk about the physical meaning before choosing some right reference point. So plus a constant in your potential has no affection.
The second thing is the Poisson equation is a differential equation, so what you are looking for is the potential if given the density distribution. Of course, you can treat it as a mathematical equation so you can pick up any mathematical function "V", but there is no correspondence between that and physical electron distribution.
Hope it helps.
Answered by Jack on March 9, 2021
The point to note is that when the RHS=0 you obtain the Lapalace Eq. In that case the potential V will not have any local maxima. That is the average value of V about any point cannot be lower or higher than V at that point. When $rho$ is not zero there are local maxima or minima at those ponits. The LHS is simply the curvature of V.
Answered by SAKhan on March 9, 2021
Your 1-dimensiomal Poisson equation is an inhomogeneous linear ordinary differential equation of second order for the potential $V(x)$. For the specification of a unique solution for $V(x)$, you have to specify the inhomogeneity (charge density) $rho(x)$ and two boundary conditions for the potential. Then you obtain the unique potential solution by adding two independent solutions of the homogeneous equation ($rho=0$) to one solution of the inhomogeneous equation so that the two boundary conditions are fulfilled. The two potential solutions you mention are both solutions to the homogeneous equation and thus correspond only to the case where the charge density is zero ($rho=0$).
Answered by freecharly on March 9, 2021
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